POJ2663,2506 Tri Tiling 递推 | 压缩DP

　　题目链接：http://poj.org/problem?id=2663    http://poj.org/problem?id=2506

　　简单的递推题，直接递推过去就可以了。

POJ-2663:

//STATUS:C++_AC_0MS_172KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=20;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int f[2][N];
int n;

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k,p;
    while(~scanf("%d",&n) && n!=-1)
    {
        mem(f,0);
        f[0][0]=1;p=1;
        for(i=0;i<n;i++){
            for(j=0;j<3;j++,mem(f[p=!p],0)){
                for(k=0;k<8;k++){
                    if(k&(1<<j)){
                        f[p][k&~(1<<j)]+=f[!p][k];
                    }
                    else {
                        f[p][k|(1<<j)]+=f[!p][k];
                        if( j<2 && !(k&(1<<(j+1))) ){
                            f[p][k|(1<<(j+1))]+=f[!p][k];
                        }
                    }
                }
            }
        }

        printf("%d\n",f[!p][0]);
    }
    return 0;
}

POJ-2605（Java大数）：

//STATUS:Java_AC_391MS_5532KB
import java.util.*;
import java.math.*;
import java.io.*;
import java.text.*;

public class Main {
    public static void main(String args[]){
        Scanner cin = new Scanner (new BufferedInputStream(System.in));
        int i,n;
        BigInteger a,b,c=BigInteger.valueOf(1);
        while(cin.hasNext()){
            n=cin.nextInt();
            a=BigInteger.valueOf(1);
            b=BigInteger.valueOf(3);
            if(n==1)c=BigInteger.valueOf(1);
            if(n==2)c=BigInteger.valueOf(3);
            for(i=3;i<=n;i++){
                c=b.add(a.multiply(BigInteger.valueOf(2)));
                a=b;b=c;
            }
            System.out.println(c);
        }
    }
}